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How to take the beginning and end parts of a list with simpler syntax?
Calling Table with custom iteratorExpectation over a list with nested-mapped expressionsDeleting powers of a common prime except the highest one from a list with further only primesChange orientation of GradientOrientationFilter from high value to lowHow to return the unevaluated variable names from a list of variables which have been declaredHow can I make ListPlot use my iteration as the x coord?Build a list from applying a recursive function on another listHow to split list into segments based on first element of sublist, and remove ineligible segments
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10
$begingroup$
So basically you have a table of some values, let's call it a:
a=Table[n,n,10];
(*1, 2, 3, 4, 5, 6, 7, 8, 9, 10*)
And you want to take the beginning and end parts of the table, within the same line, and do something with them. It is my intention to change them all to 0, but really you could change them to some other variable, say b. We want the parts of the list that do not include some inner values, we'll do 4,5,6 which happen to correspond to the indexes 4 through 6, convenient, no?
a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]] = b; a
(*b, b, b, 4, 5, 6, b, b, b, b*)
What's a simpler syntax for this? How can we call the end and beginning values of a list in a simpler manner?
a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]]
(*1, 2, 3, 7, 8, 9, 10*)
The ideal input would be, to me, something like
a[[1;;3;7;;10]]
(*1, 2, 3, 7, 8, 9, 10*)
But this gives
(*7, 8, 9, 10*)
As expected.
How can we simplify the syntax to call the beginning and end parts of a list, leaving out some chosen middle portion?
list-manipulation syntax
asked Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
$endgroup$
|
show 6 more comments
10
$begingroup$
So basically you have a table of some values, let's call it a:
a=Table[n,n,10];
(*1, 2, 3, 4, 5, 6, 7, 8, 9, 10*)
And you want to take the beginning and end parts of the table, within the same line, and do something with them. It is my intention to change them all to 0, but really you could change them to some other variable, say b. We want the parts of the list that do not include some inner values, we'll do 4,5,6 which happen to correspond to the indexes 4 through 6, convenient, no?
a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]] = b; a
(*b, b, b, 4, 5, 6, b, b, b, b*)
What's a simpler syntax for this? How can we call the end and beginning values of a list in a simpler manner?
a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]]
(*1, 2, 3, 7, 8, 9, 10*)
The ideal input would be, to me, something like
a[[1;;3;7;;10]]
(*1, 2, 3, 7, 8, 9, 10*)
But this gives
(*7, 8, 9, 10*)
As expected.
How can we simplify the syntax to call the beginning and end parts of a list, leaving out some chosen middle portion?
list-manipulation syntax
asked Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
$endgroup$
1
$begingroup$
Select[ a, ! MemberQ[ 4, 5, 6, #] & ]
$endgroup$
– LouisB
Aug 11 at 19:23
3
$begingroup$
Complement[ a, 4, 5, 6 ]
$endgroup$
– LouisB
Aug 11 at 20:02
3
$begingroup$
Cheating a little, we could write(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.
$endgroup$
– WReach
Aug 11 at 20:39
1
$begingroup$
If it's the values, then inTable[n^2,n,10]the initial segment would be the span1 ;; 1and the final segment would be3 ;; 10and the excluded segment (the positions not to be changed) would be2 ;; 2, if the excluded values were 4, 5, 6 -- no?
$endgroup$
– Michael E2
Aug 11 at 22:48
1
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
Aug 11 at 23:10
|
show 6 more comments
10
10
10
2
$begingroup$
So basically you have a table of some values, let's call it a:
a=Table[n,n,10];
(*1, 2, 3, 4, 5, 6, 7, 8, 9, 10*)
And you want to take the beginning and end parts of the table, within the same line, and do something with them. It is my intention to change them all to 0, but really you could change them to some other variable, say b. We want the parts of the list that do not include some inner values, we'll do 4,5,6 which happen to correspond to the indexes 4 through 6, convenient, no?
a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]] = b; a
(*b, b, b, 4, 5, 6, b, b, b, b*)
What's a simpler syntax for this? How can we call the end and beginning values of a list in a simpler manner?
a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]]
(*1, 2, 3, 7, 8, 9, 10*)
The ideal input would be, to me, something like
a[[1;;3;7;;10]]
(*1, 2, 3, 7, 8, 9, 10*)
But this gives
(*7, 8, 9, 10*)
As expected.
How can we simplify the syntax to call the beginning and end parts of a list, leaving out some chosen middle portion?
list-manipulation syntax
asked Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
$endgroup$
So basically you have a table of some values, let's call it a:
a=Table[n,n,10];
(*1, 2, 3, 4, 5, 6, 7, 8, 9, 10*)
And you want to take the beginning and end parts of the table, within the same line, and do something with them. It is my intention to change them all to 0, but really you could change them to some other variable, say b. We want the parts of the list that do not include some inner values, we'll do 4,5,6 which happen to correspond to the indexes 4 through 6, convenient, no?
a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]] = b; a
(*b, b, b, 4, 5, 6, b, b, b, b*)
What's a simpler syntax for this? How can we call the end and beginning values of a list in a simpler manner?
a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]]
(*1, 2, 3, 7, 8, 9, 10*)
The ideal input would be, to me, something like
a[[1;;3;7;;10]]
(*1, 2, 3, 7, 8, 9, 10*)
But this gives
(*7, 8, 9, 10*)
As expected.
How can we simplify the syntax to call the beginning and end parts of a list, leaving out some chosen middle portion?
list-manipulation syntax
list-manipulation syntax
asked Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
asked Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
asked Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
asked Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
asked Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
8611 gold badge2 silver badges17 bronze badges
1
$begingroup$
Select[ a, ! MemberQ[ 4, 5, 6, #] & ]
$endgroup$
– LouisB
Aug 11 at 19:23
3
$begingroup$
Complement[ a, 4, 5, 6 ]
$endgroup$
– LouisB
Aug 11 at 20:02
3
$begingroup$
Cheating a little, we could write(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.
$endgroup$
– WReach
Aug 11 at 20:39
1
$begingroup$
If it's the values, then inTable[n^2,n,10]the initial segment would be the span1 ;; 1and the final segment would be3 ;; 10and the excluded segment (the positions not to be changed) would be2 ;; 2, if the excluded values were 4, 5, 6 -- no?
$endgroup$
– Michael E2
Aug 11 at 22:48
1
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
Aug 11 at 23:10
|
show 6 more comments
1
$begingroup$
Select[ a, ! MemberQ[ 4, 5, 6, #] & ]
$endgroup$
– LouisB
Aug 11 at 19:23
3
$begingroup$
Complement[ a, 4, 5, 6 ]
$endgroup$
– LouisB
Aug 11 at 20:02
3
$begingroup$
Cheating a little, we could write(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.
$endgroup$
– WReach
Aug 11 at 20:39
1
$begingroup$
If it's the values, then inTable[n^2,n,10]the initial segment would be the span1 ;; 1and the final segment would be3 ;; 10and the excluded segment (the positions not to be changed) would be2 ;; 2, if the excluded values were 4, 5, 6 -- no?
$endgroup$
– Michael E2
Aug 11 at 22:48
1
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
Aug 11 at 23:10
1
1
$begingroup$
Select[ a, ! MemberQ[ 4, 5, 6, #] & ]$endgroup$
– LouisB
Aug 11 at 19:23
$begingroup$
Select[ a, ! MemberQ[ 4, 5, 6, #] & ]$endgroup$
– LouisB
Aug 11 at 19:23
3
3
$begingroup$
Complement[ a, 4, 5, 6 ]$endgroup$
– LouisB
Aug 11 at 20:02
$begingroup$
Complement[ a, 4, 5, 6 ]$endgroup$
– LouisB
Aug 11 at 20:02
3
3
$begingroup$
Cheating a little, we could write
(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.$endgroup$
– WReach
Aug 11 at 20:39
$begingroup$
Cheating a little, we could write
(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.$endgroup$
– WReach
Aug 11 at 20:39
1
1
$begingroup$
If it's the values, then in
Table[n^2,n,10] the initial segment would be the span 1 ;; 1 and the final segment would be 3 ;; 10 and the excluded segment (the positions not to be changed) would be 2 ;; 2, if the excluded values were 4, 5, 6 -- no?$endgroup$
– Michael E2
Aug 11 at 22:48
$begingroup$
If it's the values, then in
Table[n^2,n,10] the initial segment would be the span 1 ;; 1 and the final segment would be 3 ;; 10 and the excluded segment (the positions not to be changed) would be 2 ;; 2, if the excluded values were 4, 5, 6 -- no?$endgroup$
– Michael E2
Aug 11 at 22:48
1
1
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
Aug 11 at 23:10
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
Aug 11 at 23:10
|
show 6 more comments
7 Answers
7
active
oldest
votes
9
$begingroup$
Update: You can also try MapAt:
a = Range[10];
a = MapAt[b &, a, ;; 3, 7 ;;]
b, b, b, 4, 5, 6, b, b, b, b
Or ReplaceAll
a = Range[10];
a /. Alternatives @@ Drop[a, 4;;6] -> b
b, b, b, 4, 5, 6, b, b, b, b
Original answer: Try Drop:
Drop[Range @ 10, 4, 6]
1, 2, 3, 7, 8, 9, 10
Drop[Range @ 10, 4 ;; 6]
1, 2, 3, 7, 8, 9, 10
Drop[CharacterRange["a", "j"], 4, 6]
"a", "b", "c", "g", "h", "i", "j"
answered Aug 11 at 18:50
kglrkglr
226k10 gold badges256 silver badges515 bronze badges
$endgroup$
1
$begingroup$
Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we doa[[Drop[Range@10, 4, 6]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
$endgroup$
– CA Trevillian
Aug 11 at 19:01
$begingroup$
I've accepted your answer based on these parameters: generality of implementation, multiple versions, & the number of votes. Thank you!
$endgroup$
– CA Trevillian
Aug 17 at 2:29
add a comment
|
7
$begingroup$
If it is okay to perform two assignments instead of one, then we can write:
(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10
Scan is probably better since it does not bother constructing the result list that we are just going to discard anyway:
Scan[(a[[#]] = b) &, 1 ;; 3, 7 ;; 10]
answered Aug 11 at 20:52
WReachWReach
56.5k2 gold badges125 silver badges224 bronze badges
$endgroup$
$begingroup$
Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one,Scanneeds more use these days, imo.
$endgroup$
– CA Trevillian
Aug 11 at 20:59
add a comment
|
6
$begingroup$
Perhaps this?:
ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
(* b, b, b, 4, 5, 6, b, b, b, b *)
answered Aug 11 at 23:09
Michael E2Michael E2
163k13 gold badges221 silver badges523 bronze badges
$endgroup$
$begingroup$
Wow nice!! I like the use ofRuleDelayed, am I correct in understanding that you could use eitherTable[n,n,10]orTable[n^2,n,10]and we should see the 4th-6th indexed values preserved, asReplacePartreplaces the ith index? The use ofNotprovides a good deal of readability as well.
$endgroup$
– CA Trevillian
Aug 11 at 23:20
2
$begingroup$
@CATrevillian Yes, it depends only on the index (part/position), not the value.
$endgroup$
– Michael E2
Aug 11 at 23:23
add a comment
|
5
$begingroup$
By far the simplest way is to do it in two lines:
a = Table[RandomReal[], n, 10];
a[[1 ;; 3]] = b;
a[[7 ;; 10]] = b;
a
This gives the output:
b, b, b, 0.378846, 0.475894, 0.533768, b, b, b, b
answered Aug 12 at 1:54
axsvl77axsvl77
8715 silver badges16 bronze badges
$endgroup$
add a comment
|
4
$begingroup$
We can use Union[] to perform this operation in a simpler manner:
a[[Range[1, 3] [Union] Range[7, 10]]]
(*1, 2, 3, 7, 8, 9, 10*)
But this is still decently messy, and the readability is not too high. Perhaps someone else has a better method?
a[[Range[1, 3] [Union] Range[7, 10]]] = b; a
(*b, b, b, 4, 5, 6, b, b, b, b*)
answered Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
$endgroup$
add a comment
|
3
$begingroup$
For me, the least fussy approach involves using ArrayPad[] twice:
ArrayPad[ArrayPad[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, -3, -4], 3, 4, b]
b, b, b, 4, 5, 6, b, b, b, b
answered Aug 12 at 9:07
J. M. will be back soon♦J. M. will be back soon
103k11 gold badges320 silver badges479 bronze badges
$endgroup$
$begingroup$
Can this be used to rip out the middle portion and present just the edge parts? I suspect so, but I don’t understand this well enough on the surface. Thanks in advance! Very nice usage :)
$endgroup$
– CA Trevillian
Aug 12 at 13:07
1
$begingroup$
"rip out the middle portion and present just the edge parts" - for that case, I would be usingTake[]/Part[]instead.ArrayPad[]is designed to act on both ends of a list, either by adding or removing elements (depending on the signs of the integers used in the second argument).
$endgroup$
– J. M. will be back soon♦
Aug 15 at 5:06
add a comment
|
3
$begingroup$
How about
Range[10] // #[[;; 3]]~Join~#[[7 ;;]] &
answered Aug 12 at 17:56
user2757771user2757771
1463 bronze badges
$endgroup$
add a comment
|
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
9
$begingroup$
Update: You can also try MapAt:
a = Range[10];
a = MapAt[b &, a, ;; 3, 7 ;;]
b, b, b, 4, 5, 6, b, b, b, b
Or ReplaceAll
a = Range[10];
a /. Alternatives @@ Drop[a, 4;;6] -> b
b, b, b, 4, 5, 6, b, b, b, b
Original answer: Try Drop:
Drop[Range @ 10, 4, 6]
1, 2, 3, 7, 8, 9, 10
Drop[Range @ 10, 4 ;; 6]
1, 2, 3, 7, 8, 9, 10
Drop[CharacterRange["a", "j"], 4, 6]
"a", "b", "c", "g", "h", "i", "j"
answered Aug 11 at 18:50
kglrkglr
226k10 gold badges256 silver badges515 bronze badges
$endgroup$
1
$begingroup$
Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we doa[[Drop[Range@10, 4, 6]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
$endgroup$
– CA Trevillian
Aug 11 at 19:01
$begingroup$
I've accepted your answer based on these parameters: generality of implementation, multiple versions, & the number of votes. Thank you!
$endgroup$
– CA Trevillian
Aug 17 at 2:29
add a comment
|
9
$begingroup$
Update: You can also try MapAt:
a = Range[10];
a = MapAt[b &, a, ;; 3, 7 ;;]
b, b, b, 4, 5, 6, b, b, b, b
Or ReplaceAll
a = Range[10];
a /. Alternatives @@ Drop[a, 4;;6] -> b
b, b, b, 4, 5, 6, b, b, b, b
Original answer: Try Drop:
Drop[Range @ 10, 4, 6]
1, 2, 3, 7, 8, 9, 10
Drop[Range @ 10, 4 ;; 6]
1, 2, 3, 7, 8, 9, 10
Drop[CharacterRange["a", "j"], 4, 6]
"a", "b", "c", "g", "h", "i", "j"
answered Aug 11 at 18:50
kglrkglr
226k10 gold badges256 silver badges515 bronze badges
$endgroup$
1
$begingroup$
Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we doa[[Drop[Range@10, 4, 6]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
$endgroup$
– CA Trevillian
Aug 11 at 19:01
$begingroup$
I've accepted your answer based on these parameters: generality of implementation, multiple versions, & the number of votes. Thank you!
$endgroup$
– CA Trevillian
Aug 17 at 2:29
add a comment
|
9
9
9
$begingroup$
Update: You can also try MapAt:
a = Range[10];
a = MapAt[b &, a, ;; 3, 7 ;;]
b, b, b, 4, 5, 6, b, b, b, b
Or ReplaceAll
a = Range[10];
a /. Alternatives @@ Drop[a, 4;;6] -> b
b, b, b, 4, 5, 6, b, b, b, b
Original answer: Try Drop:
Drop[Range @ 10, 4, 6]
1, 2, 3, 7, 8, 9, 10
Drop[Range @ 10, 4 ;; 6]
1, 2, 3, 7, 8, 9, 10
Drop[CharacterRange["a", "j"], 4, 6]
"a", "b", "c", "g", "h", "i", "j"
answered Aug 11 at 18:50
kglrkglr
226k10 gold badges256 silver badges515 bronze badges
$endgroup$
Update: You can also try MapAt:
a = Range[10];
a = MapAt[b &, a, ;; 3, 7 ;;]
b, b, b, 4, 5, 6, b, b, b, b
Or ReplaceAll
a = Range[10];
a /. Alternatives @@ Drop[a, 4;;6] -> b
b, b, b, 4, 5, 6, b, b, b, b
Original answer: Try Drop:
Drop[Range @ 10, 4, 6]
1, 2, 3, 7, 8, 9, 10
Drop[Range @ 10, 4 ;; 6]
1, 2, 3, 7, 8, 9, 10
Drop[CharacterRange["a", "j"], 4, 6]
"a", "b", "c", "g", "h", "i", "j"
answered Aug 11 at 18:50
kglrkglr
226k10 gold badges256 silver badges515 bronze badges
edited Aug 12 at 4:52
answered Aug 11 at 18:50
kglrkglr
226k10 gold badges256 silver badges515 bronze badges
answered Aug 11 at 18:50
kglrkglr
226k10 gold badges256 silver badges515 bronze badges
answered Aug 11 at 18:50
kglrkglr
226k10 gold badges256 silver badges515 bronze badges
226k10 gold badges256 silver badges515 bronze badges
1
$begingroup$
Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we doa[[Drop[Range@10, 4, 6]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
$endgroup$
– CA Trevillian
Aug 11 at 19:01
$begingroup$
I've accepted your answer based on these parameters: generality of implementation, multiple versions, & the number of votes. Thank you!
$endgroup$
– CA Trevillian
Aug 17 at 2:29
add a comment
|
1
$begingroup$
Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we doa[[Drop[Range@10, 4, 6]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
$endgroup$
– CA Trevillian
Aug 11 at 19:01
$begingroup$
I've accepted your answer based on these parameters: generality of implementation, multiple versions, & the number of votes. Thank you!
$endgroup$
– CA Trevillian
Aug 17 at 2:29
1
1
$begingroup$
Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we do
a[[Drop[Range@10, 4, 6]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1$endgroup$
– CA Trevillian
Aug 11 at 19:01
$begingroup$
Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we do
a[[Drop[Range@10, 4, 6]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1$endgroup$
– CA Trevillian
Aug 11 at 19:01
$begingroup$
I've accepted your answer based on these parameters: generality of implementation, multiple versions, & the number of votes. Thank you!
$endgroup$
– CA Trevillian
Aug 17 at 2:29
$begingroup$
I've accepted your answer based on these parameters: generality of implementation, multiple versions, & the number of votes. Thank you!
$endgroup$
– CA Trevillian
Aug 17 at 2:29
add a comment
|
7
$begingroup$
If it is okay to perform two assignments instead of one, then we can write:
(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10
Scan is probably better since it does not bother constructing the result list that we are just going to discard anyway:
Scan[(a[[#]] = b) &, 1 ;; 3, 7 ;; 10]
answered Aug 11 at 20:52
WReachWReach
56.5k2 gold badges125 silver badges224 bronze badges
$endgroup$
$begingroup$
Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one,Scanneeds more use these days, imo.
$endgroup$
– CA Trevillian
Aug 11 at 20:59
add a comment
|
7
$begingroup$
If it is okay to perform two assignments instead of one, then we can write:
(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10
Scan is probably better since it does not bother constructing the result list that we are just going to discard anyway:
Scan[(a[[#]] = b) &, 1 ;; 3, 7 ;; 10]
answered Aug 11 at 20:52
WReachWReach
56.5k2 gold badges125 silver badges224 bronze badges
$endgroup$
$begingroup$
Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one,Scanneeds more use these days, imo.
$endgroup$
– CA Trevillian
Aug 11 at 20:59
add a comment
|
7
7
7
$begingroup$
If it is okay to perform two assignments instead of one, then we can write:
(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10
Scan is probably better since it does not bother constructing the result list that we are just going to discard anyway:
Scan[(a[[#]] = b) &, 1 ;; 3, 7 ;; 10]
answered Aug 11 at 20:52
WReachWReach
56.5k2 gold badges125 silver badges224 bronze badges
$endgroup$
If it is okay to perform two assignments instead of one, then we can write:
(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10
Scan is probably better since it does not bother constructing the result list that we are just going to discard anyway:
Scan[(a[[#]] = b) &, 1 ;; 3, 7 ;; 10]
answered Aug 11 at 20:52
WReachWReach
56.5k2 gold badges125 silver badges224 bronze badges
answered Aug 11 at 20:52
WReachWReach
56.5k2 gold badges125 silver badges224 bronze badges
answered Aug 11 at 20:52
WReachWReach
56.5k2 gold badges125 silver badges224 bronze badges
answered Aug 11 at 20:52
WReachWReach
56.5k2 gold badges125 silver badges224 bronze badges
56.5k2 gold badges125 silver badges224 bronze badges
$begingroup$
Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one,Scanneeds more use these days, imo.
$endgroup$
– CA Trevillian
Aug 11 at 20:59
add a comment
|
$begingroup$
Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one,Scanneeds more use these days, imo.
$endgroup$
– CA Trevillian
Aug 11 at 20:59
$begingroup$
Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one,
Scan needs more use these days, imo.$endgroup$
– CA Trevillian
Aug 11 at 20:59
$begingroup$
Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one,
Scan needs more use these days, imo.$endgroup$
– CA Trevillian
Aug 11 at 20:59
add a comment
|
6
$begingroup$
Perhaps this?:
ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
(* b, b, b, 4, 5, 6, b, b, b, b *)
answered Aug 11 at 23:09
Michael E2Michael E2
163k13 gold badges221 silver badges523 bronze badges
$endgroup$
$begingroup$
Wow nice!! I like the use ofRuleDelayed, am I correct in understanding that you could use eitherTable[n,n,10]orTable[n^2,n,10]and we should see the 4th-6th indexed values preserved, asReplacePartreplaces the ith index? The use ofNotprovides a good deal of readability as well.
$endgroup$
– CA Trevillian
Aug 11 at 23:20
2
$begingroup$
@CATrevillian Yes, it depends only on the index (part/position), not the value.
$endgroup$
– Michael E2
Aug 11 at 23:23
add a comment
|
6
$begingroup$
Perhaps this?:
ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
(* b, b, b, 4, 5, 6, b, b, b, b *)
answered Aug 11 at 23:09
Michael E2Michael E2
163k13 gold badges221 silver badges523 bronze badges
$endgroup$
$begingroup$
Wow nice!! I like the use ofRuleDelayed, am I correct in understanding that you could use eitherTable[n,n,10]orTable[n^2,n,10]and we should see the 4th-6th indexed values preserved, asReplacePartreplaces the ith index? The use ofNotprovides a good deal of readability as well.
$endgroup$
– CA Trevillian
Aug 11 at 23:20
2
$begingroup$
@CATrevillian Yes, it depends only on the index (part/position), not the value.
$endgroup$
– Michael E2
Aug 11 at 23:23
add a comment
|
6
6
6
$begingroup$
Perhaps this?:
ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
(* b, b, b, 4, 5, 6, b, b, b, b *)
answered Aug 11 at 23:09
Michael E2Michael E2
163k13 gold badges221 silver badges523 bronze badges
$endgroup$
Perhaps this?:
ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
(* b, b, b, 4, 5, 6, b, b, b, b *)
answered Aug 11 at 23:09
Michael E2Michael E2
163k13 gold badges221 silver badges523 bronze badges
answered Aug 11 at 23:09
Michael E2Michael E2
163k13 gold badges221 silver badges523 bronze badges
answered Aug 11 at 23:09
Michael E2Michael E2
163k13 gold badges221 silver badges523 bronze badges
answered Aug 11 at 23:09
Michael E2Michael E2
163k13 gold badges221 silver badges523 bronze badges
163k13 gold badges221 silver badges523 bronze badges
$begingroup$
Wow nice!! I like the use ofRuleDelayed, am I correct in understanding that you could use eitherTable[n,n,10]orTable[n^2,n,10]and we should see the 4th-6th indexed values preserved, asReplacePartreplaces the ith index? The use ofNotprovides a good deal of readability as well.
$endgroup$
– CA Trevillian
Aug 11 at 23:20
2
$begingroup$
@CATrevillian Yes, it depends only on the index (part/position), not the value.
$endgroup$
– Michael E2
Aug 11 at 23:23
add a comment
|
$begingroup$
Wow nice!! I like the use ofRuleDelayed, am I correct in understanding that you could use eitherTable[n,n,10]orTable[n^2,n,10]and we should see the 4th-6th indexed values preserved, asReplacePartreplaces the ith index? The use ofNotprovides a good deal of readability as well.
$endgroup$
– CA Trevillian
Aug 11 at 23:20
2
$begingroup$
@CATrevillian Yes, it depends only on the index (part/position), not the value.
$endgroup$
– Michael E2
Aug 11 at 23:23
$begingroup$
Wow nice!! I like the use of
RuleDelayed, am I correct in understanding that you could use either Table[n,n,10] or Table[n^2,n,10] and we should see the 4th-6th indexed values preserved, as ReplacePart replaces the ith index? The use of Not provides a good deal of readability as well.$endgroup$
– CA Trevillian
Aug 11 at 23:20
$begingroup$
Wow nice!! I like the use of
RuleDelayed, am I correct in understanding that you could use either Table[n,n,10] or Table[n^2,n,10] and we should see the 4th-6th indexed values preserved, as ReplacePart replaces the ith index? The use of Not provides a good deal of readability as well.$endgroup$
– CA Trevillian
Aug 11 at 23:20
2
2
$begingroup$
@CATrevillian Yes, it depends only on the index (part/position), not the value.
$endgroup$
– Michael E2
Aug 11 at 23:23
$begingroup$
@CATrevillian Yes, it depends only on the index (part/position), not the value.
$endgroup$
– Michael E2
Aug 11 at 23:23
add a comment
|
5
$begingroup$
By far the simplest way is to do it in two lines:
a = Table[RandomReal[], n, 10];
a[[1 ;; 3]] = b;
a[[7 ;; 10]] = b;
a
This gives the output:
b, b, b, 0.378846, 0.475894, 0.533768, b, b, b, b
answered Aug 12 at 1:54
axsvl77axsvl77
8715 silver badges16 bronze badges
$endgroup$
add a comment
|
5
$begingroup$
By far the simplest way is to do it in two lines:
a = Table[RandomReal[], n, 10];
a[[1 ;; 3]] = b;
a[[7 ;; 10]] = b;
a
This gives the output:
b, b, b, 0.378846, 0.475894, 0.533768, b, b, b, b
answered Aug 12 at 1:54
axsvl77axsvl77
8715 silver badges16 bronze badges
$endgroup$
add a comment
|
5
5
5
$begingroup$
By far the simplest way is to do it in two lines:
a = Table[RandomReal[], n, 10];
a[[1 ;; 3]] = b;
a[[7 ;; 10]] = b;
a
This gives the output:
b, b, b, 0.378846, 0.475894, 0.533768, b, b, b, b
answered Aug 12 at 1:54
axsvl77axsvl77
8715 silver badges16 bronze badges
$endgroup$
By far the simplest way is to do it in two lines:
a = Table[RandomReal[], n, 10];
a[[1 ;; 3]] = b;
a[[7 ;; 10]] = b;
a
This gives the output:
b, b, b, 0.378846, 0.475894, 0.533768, b, b, b, b
answered Aug 12 at 1:54
axsvl77axsvl77
8715 silver badges16 bronze badges
edited Aug 12 at 5:04
answered Aug 12 at 1:54
axsvl77axsvl77
8715 silver badges16 bronze badges
answered Aug 12 at 1:54
axsvl77axsvl77
8715 silver badges16 bronze badges
answered Aug 12 at 1:54
axsvl77axsvl77
8715 silver badges16 bronze badges
8715 silver badges16 bronze badges
add a comment
|
add a comment
|
4
$begingroup$
We can use Union[] to perform this operation in a simpler manner:
a[[Range[1, 3] [Union] Range[7, 10]]]
(*1, 2, 3, 7, 8, 9, 10*)
But this is still decently messy, and the readability is not too high. Perhaps someone else has a better method?
a[[Range[1, 3] [Union] Range[7, 10]]] = b; a
(*b, b, b, 4, 5, 6, b, b, b, b*)
answered Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
$endgroup$
add a comment
|
4
$begingroup$
We can use Union[] to perform this operation in a simpler manner:
a[[Range[1, 3] [Union] Range[7, 10]]]
(*1, 2, 3, 7, 8, 9, 10*)
But this is still decently messy, and the readability is not too high. Perhaps someone else has a better method?
a[[Range[1, 3] [Union] Range[7, 10]]] = b; a
(*b, b, b, 4, 5, 6, b, b, b, b*)
answered Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
$endgroup$
add a comment
|
4
4
4
$begingroup$
We can use Union[] to perform this operation in a simpler manner:
a[[Range[1, 3] [Union] Range[7, 10]]]
(*1, 2, 3, 7, 8, 9, 10*)
But this is still decently messy, and the readability is not too high. Perhaps someone else has a better method?
a[[Range[1, 3] [Union] Range[7, 10]]] = b; a
(*b, b, b, 4, 5, 6, b, b, b, b*)
answered Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
$endgroup$
We can use Union[] to perform this operation in a simpler manner:
a[[Range[1, 3] [Union] Range[7, 10]]]
(*1, 2, 3, 7, 8, 9, 10*)
But this is still decently messy, and the readability is not too high. Perhaps someone else has a better method?
a[[Range[1, 3] [Union] Range[7, 10]]] = b; a
(*b, b, b, 4, 5, 6, b, b, b, b*)
answered Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
answered Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
answered Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
answered Aug 11 at 18:42
CA TrevillianCA Trevillian
8611 gold badge2 silver badges17 bronze badges
8611 gold badge2 silver badges17 bronze badges
add a comment
|
add a comment
|
3
$begingroup$
For me, the least fussy approach involves using ArrayPad[] twice:
ArrayPad[ArrayPad[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, -3, -4], 3, 4, b]
b, b, b, 4, 5, 6, b, b, b, b
answered Aug 12 at 9:07
J. M. will be back soon♦J. M. will be back soon
103k11 gold badges320 silver badges479 bronze badges
$endgroup$
$begingroup$
Can this be used to rip out the middle portion and present just the edge parts? I suspect so, but I don’t understand this well enough on the surface. Thanks in advance! Very nice usage :)
$endgroup$
– CA Trevillian
Aug 12 at 13:07
1
$begingroup$
"rip out the middle portion and present just the edge parts" - for that case, I would be usingTake[]/Part[]instead.ArrayPad[]is designed to act on both ends of a list, either by adding or removing elements (depending on the signs of the integers used in the second argument).
$endgroup$
– J. M. will be back soon♦
Aug 15 at 5:06
add a comment
|
3
$begingroup$
For me, the least fussy approach involves using ArrayPad[] twice:
ArrayPad[ArrayPad[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, -3, -4], 3, 4, b]
b, b, b, 4, 5, 6, b, b, b, b
answered Aug 12 at 9:07
J. M. will be back soon♦J. M. will be back soon
103k11 gold badges320 silver badges479 bronze badges
$endgroup$
$begingroup$
Can this be used to rip out the middle portion and present just the edge parts? I suspect so, but I don’t understand this well enough on the surface. Thanks in advance! Very nice usage :)
$endgroup$
– CA Trevillian
Aug 12 at 13:07
1
$begingroup$
"rip out the middle portion and present just the edge parts" - for that case, I would be usingTake[]/Part[]instead.ArrayPad[]is designed to act on both ends of a list, either by adding or removing elements (depending on the signs of the integers used in the second argument).
$endgroup$
– J. M. will be back soon♦
Aug 15 at 5:06
add a comment
|
3
3
3
$begingroup$
For me, the least fussy approach involves using ArrayPad[] twice:
ArrayPad[ArrayPad[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, -3, -4], 3, 4, b]
b, b, b, 4, 5, 6, b, b, b, b
answered Aug 12 at 9:07
J. M. will be back soon♦J. M. will be back soon
103k11 gold badges320 silver badges479 bronze badges
$endgroup$
For me, the least fussy approach involves using ArrayPad[] twice:
ArrayPad[ArrayPad[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, -3, -4], 3, 4, b]
b, b, b, 4, 5, 6, b, b, b, b
answered Aug 12 at 9:07
J. M. will be back soon♦J. M. will be back soon
103k11 gold badges320 silver badges479 bronze badges
answered Aug 12 at 9:07
J. M. will be back soon♦J. M. will be back soon
103k11 gold badges320 silver badges479 bronze badges
answered Aug 12 at 9:07
J. M. will be back soon♦J. M. will be back soon
103k11 gold badges320 silver badges479 bronze badges
answered Aug 12 at 9:07
J. M. will be back soon♦J. M. will be back soon
103k11 gold badges320 silver badges479 bronze badges
103k11 gold badges320 silver badges479 bronze badges
$begingroup$
Can this be used to rip out the middle portion and present just the edge parts? I suspect so, but I don’t understand this well enough on the surface. Thanks in advance! Very nice usage :)
$endgroup$
– CA Trevillian
Aug 12 at 13:07
1
$begingroup$
"rip out the middle portion and present just the edge parts" - for that case, I would be usingTake[]/Part[]instead.ArrayPad[]is designed to act on both ends of a list, either by adding or removing elements (depending on the signs of the integers used in the second argument).
$endgroup$
– J. M. will be back soon♦
Aug 15 at 5:06
add a comment
|
$begingroup$
Can this be used to rip out the middle portion and present just the edge parts? I suspect so, but I don’t understand this well enough on the surface. Thanks in advance! Very nice usage :)
$endgroup$
– CA Trevillian
Aug 12 at 13:07
1
$begingroup$
"rip out the middle portion and present just the edge parts" - for that case, I would be usingTake[]/Part[]instead.ArrayPad[]is designed to act on both ends of a list, either by adding or removing elements (depending on the signs of the integers used in the second argument).
$endgroup$
– J. M. will be back soon♦
Aug 15 at 5:06
$begingroup$
Can this be used to rip out the middle portion and present just the edge parts? I suspect so, but I don’t understand this well enough on the surface. Thanks in advance! Very nice usage :)
$endgroup$
– CA Trevillian
Aug 12 at 13:07
$begingroup$
Can this be used to rip out the middle portion and present just the edge parts? I suspect so, but I don’t understand this well enough on the surface. Thanks in advance! Very nice usage :)
$endgroup$
– CA Trevillian
Aug 12 at 13:07
1
1
$begingroup$
"rip out the middle portion and present just the edge parts" - for that case, I would be using
Take[]/Part[] instead. ArrayPad[] is designed to act on both ends of a list, either by adding or removing elements (depending on the signs of the integers used in the second argument).$endgroup$
– J. M. will be back soon♦
Aug 15 at 5:06
$begingroup$
"rip out the middle portion and present just the edge parts" - for that case, I would be using
Take[]/Part[] instead. ArrayPad[] is designed to act on both ends of a list, either by adding or removing elements (depending on the signs of the integers used in the second argument).$endgroup$
– J. M. will be back soon♦
Aug 15 at 5:06
add a comment
|
3
$begingroup$
How about
Range[10] // #[[;; 3]]~Join~#[[7 ;;]] &
answered Aug 12 at 17:56
user2757771user2757771
1463 bronze badges
$endgroup$
add a comment
|
3
$begingroup$
How about
Range[10] // #[[;; 3]]~Join~#[[7 ;;]] &
answered Aug 12 at 17:56
user2757771user2757771
1463 bronze badges
$endgroup$
add a comment
|
3
3
3
$begingroup$
How about
Range[10] // #[[;; 3]]~Join~#[[7 ;;]] &
answered Aug 12 at 17:56
user2757771user2757771
1463 bronze badges
$endgroup$
How about
Range[10] // #[[;; 3]]~Join~#[[7 ;;]] &
answered Aug 12 at 17:56
user2757771user2757771
1463 bronze badges
answered Aug 12 at 17:56
user2757771user2757771
1463 bronze badges
answered Aug 12 at 17:56
user2757771user2757771
1463 bronze badges
answered Aug 12 at 17:56
user2757771user2757771
1463 bronze badges
1463 bronze badges
add a comment
|
add a comment
|
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1
$begingroup$
Select[ a, ! MemberQ[ 4, 5, 6, #] & ]$endgroup$
– LouisB
Aug 11 at 19:23
3
$begingroup$
Complement[ a, 4, 5, 6 ]$endgroup$
– LouisB
Aug 11 at 20:02
3
$begingroup$
Cheating a little, we could write
(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.$endgroup$
– WReach
Aug 11 at 20:39
1
$begingroup$
If it's the values, then in
Table[n^2,n,10]the initial segment would be the span1 ;; 1and the final segment would be3 ;; 10and the excluded segment (the positions not to be changed) would be2 ;; 2, if the excluded values were 4, 5, 6 -- no?$endgroup$
– Michael E2
Aug 11 at 22:48
1
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
Aug 11 at 23:10